# What is the vertex of the graph of y=x^2+3x-4?

Oct 15, 2014

This specific parabola should be transformed into the form

$y = {\left(x - h\right)}^{2} + k$

To do that, we need to perform Completing the Square on $x$

Remember that ${\left(a x + b\right)}^{2} = {a}^{2} {x}^{2} + 2 a b x + {b}^{2}$

For this problem, we have $a = 1$ and $2 a b = 3$

$\implies 2 \left(1\right) b = 3$
$\implies b = \frac{3}{2}$

Hence, for $x$ to be a perfect square, we need to add ${b}^{2}$.
However, we want both sides of the equation to remain equal, so we substract the same value again

$y = \left({x}^{2} + 3 x + {\left(\frac{3}{2}\right)}^{2}\right) - 4 - {\left(\frac{3}{2}\right)}^{2}$

Simplifying, we have
$y = {\left(x + \frac{3}{2}\right)}^{2} - \frac{25}{4}$

The parabola's vertex is at $\left(- \frac{3}{2} , \frac{25}{4}\right)$.